RI 2019 Year 5 CT Suggested Solutions

© Raffles Institution 2019 9729/J/19 2019 Y5 H2 Chemistry Term 3 Common Test – Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer B C B B D C B C C D A A D A D MCQ worked solutions 1 (Ans: B) From the mass numbers given in the table, the sample contains 10B, 11B and 23Na (23Na is the only impurity present). To compute the Ar of boron in the given sample, we need to consider only the isotopes , 10B and 11B, and their % abundance (i.e. 23Na and i ts % abundance should not be included in the calculation). Note also that the sum of % abundance of 10B and 11B is not 100%. Ar of boron in sample = ( )( ) ( )( )15.52 11 74.48 15.52   10  +     + 74.48 = 10.83 2 (Ans: C) In the redox reaction, V2O5 is reduced and Fe2+ is oxidised to Fe3+. 25VOn : en − : 2+Fen Since Oxidation: Fe2+ ⎯→ Fe3+ + e− or 2Fe2+ ⎯→ 2Fe3+ + 2e− Reduction (unbalanced): V2O5 + 2e– ⎯→ ?  Each V atom accepted 1 e–.  Oxidation no. of V decreases from +5 to +4. : 1 : 1 1 : : 2 1 : 2 : 2 Oxidation number of V 0 +2 +4 +3 A V B VO C VO2 D V2O3 3 (Ans: B) H+, being positively charged, gives  of +15o and | q m| = 1 Since particle Z gives  of –5o, particle Z is negatively charged and | q m| = 1 3 proton neutron electron q m |q m| 1 1 0 4 –3 1 3 2 1 2 2 –1 3 1 3 3 3 3 1 +2 6 1 3 4 3 3 5 –2 6 1 3 Hence, only options 2 and 4 are correct. This document is copyrighted, please do not reproduce it without permission

© Raffles Institution 2019 9729/J/19 4 (Ans: B) There is a large jump in the 4th to 5th IE for one element, and in the 6th to 7th IE for another element. Hence these two elements belong to group 14 and 16 respectively, with minimally 8 electrons. CS2 is not possible because C has only 6 electrons to be removed. Hence only SiO2 is possible. 5 (Ans: D) A (incorrect): X and Y do not have the same proton number and hence they are not isotopes. B (incorrect): Electronegativity of an element increases across the period. Hence Y has a higher electronegativity than X. C (incorrect): Since X and Y have different proton number (and hence different no. of e−), they will not have the same electronic configuration. D (correct): As the ion of Y is isoelectronic with X+, the ion of Y has a +2 charge (i.e. Y2+). Hence, the ion of Y has a higher charge density (higher charge, smaller ionic radius) than the ion of X. 6 (Ans: C) molecule structure no. of  bonds no. of σ bonds CO2 2 2 HCN 2 2 N2 2 1 NO2 1 2 (Note: the dative covalent bond is also a σ bond) 7 (Ans: B) around underlined atom molecule structure no. of bond pairs no. of lone pairs shape BCl3 3 0 trigonal planar AlCl3 3 0 trigonal planar CO2 2 0 linear SiO2 4 0 tetrahedral H2O 2 2 bent SO2 2 1 bent This document is copyrighted, please do not reproduce it without permission

© Raffles Institution 2019 9729/J/19 8 (Ans: C) A (incorrect): Polar compounds may not always be soluble in water. If the energy released from the interactions formed between the compound and water is insufficient to compensate for the energy required to overcome the compound -compound and water - water interactions, then the compound may not be soluble. For example, chloroform, CHCl3, is polar but it is insoluble in water. B (incorrect): Compounds containing polar bonds can be non-polar if the dipole moments of the polar bonds cancel each other out, resulting in zero net dipole moment. CO2 is one such example. C (correct): When a charged rod is brought near a stream of a polar compound, the polar compound is attracted to the rod because of the dipoles present on the polar compound. D (incorrect): All molecules experience id-id forces. Therefore, all polar compounds must experience both id -id and pd -pd forces. Also, some polar compounds are able to form intermolecular hydrogen bonding, in addition to id-id and pd-id interactions. 9 (Ans: C) Statement 1 (incorrect): The compound cannot form hydrogen bonds with itself because it does not have any H atoms directly bonded to an F, O or N atom. Statement 2 (correct): The compound is ionic so it can form ion -dipole interactions with water. Statement 3 (correct): |LE|  qq rr +− +−+ . Since the cation has a small charge and a very large size, the magnitude of the lattice energy is small. 10 (Ans: D) manipulation of ideal gas equation correct graph A pV = nRT since n, R and T are constants, pV = constant p  1 V hence graph is similar to y = k x B pV = nRT Since n, R and T constants, pV = constant at a given T. Hence graph is similar to x = c (vertical straight line) This document is copyrighted, please do not reproduce it without permission

© Raffles Institution 2019 9729/J/19 C pV = nRT Since n, R and p are constants, V  T Hence, graph is similar to y = kx (straight line that passes through origin) D pV = nRT  pV = RT  p = x RT M = density x RT M  density = pM RT since p, M and R are constants, density  1 T hence graph is similar to y = k x 11 (Ans: A) Assumption of an ideal gas: 1. the gas particles have negligible volume and 2. the gas particles exert negligible attractive forces on one another. Neon behaves most ideally at rtp as the • volume of neon gas particles is the smallest and • intermolecular forces of attraction (id-id) in neon is the weakest (Intermolecular hydrogen bonding in NH 3 is stronger than the id -id interaction in Ne, Cl2 and CO2. Amongst Ne, Cl2 and CO2, Ne has the weakest id-id interaction as it has the smallest and least polarisable electron cloud.) 12 (Ans: A) Statement 1 (incorrect): Standard enthalpy change of atomisation of Br 2(l) refers to the energy absorbed to form one mole of gaseous Br atoms from Br2(l) at 298 K and 1 bar. In the equation, two moles of Br(g) are formed. Statement 2 (correct): Standard enthalpy change of combustion of C(s) refers to the energy released when one mole of C(s) is burnt in excess oxygen at 298 K and 1 bar. In the equation, there is one mole of C(s) being burnt to give CO2(g). The formation of CO2(g) only indicates complete combustion. (Note: In incomplete combustion, CO(g) is formed as well). Statement 3 (incorrect): Standard enthalpy change of formation of H 2O(l) refers to the energy change when one mole of H2O(l) is formed from its constituent elements at 298 K and 1 bar. In the equation, there are two moles of H2O(l) formed. M m V m This document is copyrighted, please do not reproduce it without permission

© Raffles Institution 2019 9729/J/19 13 (Ans: D) Thermal decomposition of MgCO3 is an endothermic process as heat needs to be supplied in order for the reaction to proceed. Hence, ΔH is positive. Since the decomposition happens only at high temperatures, it means that ΔG < 0 only at high temperatures. ΔG = ΔH − TΔS Since ΔH > 0, ΔS must be positive so that at high temperatures, TΔS > ΔH, which will lead to ΔG < 0. Alternative explanation: MgCO3(s) ⎯→ MgO(s) + CO2(g) Thermal decomposition of MgCO3 is an endothermic process as heat needs to be supplied in order for the reaction to proceed. Hence, ΔH is positive. Since there is an increase in the number of gas particles after reaction , entropy has increased and the system becomes more disordered. Hence, ΔS is positive. 14 (Ans: A) By Hess’ Law: ∆Hr = − ΔHf[NH3(g)] + 3 x ΔHf[HCl(g)] + ΔHf[NCl3(l)] = − (−45.9) + 3(−92.3) + (+230.0) = −1.0 kJ mol−1 (Note: Take note of coefficients. Recall that standard enthalpy change of formation is for the formation of one mo